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Sagot :
Sure! Let's solve the equation step by step:
The given equation is:
[tex]\[2^x + \frac{1}{2^x} = 4.25\][/tex]
1. Introduce a substitution:
Let \( y = 2^x \). Our equation then becomes:
[tex]\[y + \frac{1}{y} = 4.25\][/tex]
2. Multiply both sides by \( y \):
To eliminate the fraction, multiply both sides by \( y \):
[tex]\[y^2 + 1 = 4.25y\][/tex]
3. Rearrange the equation:
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[y^2 - 4.25y + 1 = 0\][/tex]
4. Solve the quadratic equation:
We can solve this quadratic equation using the quadratic formula:
[tex]\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Where \( a = 1 \), \( b = -4.25 \), and \( c = 1 \).
5. Calculate the discriminant:
First, find the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-4.25)^2 - 4 \cdot 1 \cdot 1 = 18.0625 - 4 = 14.0625 \][/tex]
6. Find the roots:
Now plug the discriminant (\(\Delta\)) into the quadratic formula:
[tex]\[y_1 = \frac{4.25 + \sqrt{14.0625}}{2} = \frac{4.25 + 3.75}{2} = \frac{8}{2} = 4\][/tex]
[tex]\[y_2 = \frac{4.25 - \sqrt{14.0625}}{2} = \frac{4.25 - 3.75}{2} = \frac{0.5}{2} = 0.25\][/tex]
So the solutions for \( y \) are 4 and 0.25.
7. Revert the substitution:
Recall that \( y = 2^x \). Therefore, we have:
[tex]\[2^x = 4\][/tex]
[tex]\[2^x = 0.25\][/tex]
8. Solve for \( x \):
To find \( x \), we need to express each equation in terms of \( x \) using logarithms:
[tex]\[2^x = 4 \implies x = \log_2 4 = 2\][/tex]
[tex]\[2^x = 0.25 \implies x = \log_2 0.25 = -2\][/tex]
Thus, the solutions for \( x \) are \( x = 2 \) and \( x = -2 \).
In conclusion, the solution to the equation \(2^x + \frac{1}{2^x} = 4.25\) is:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
The given equation is:
[tex]\[2^x + \frac{1}{2^x} = 4.25\][/tex]
1. Introduce a substitution:
Let \( y = 2^x \). Our equation then becomes:
[tex]\[y + \frac{1}{y} = 4.25\][/tex]
2. Multiply both sides by \( y \):
To eliminate the fraction, multiply both sides by \( y \):
[tex]\[y^2 + 1 = 4.25y\][/tex]
3. Rearrange the equation:
Bring all terms to one side of the equation to form a standard quadratic equation:
[tex]\[y^2 - 4.25y + 1 = 0\][/tex]
4. Solve the quadratic equation:
We can solve this quadratic equation using the quadratic formula:
[tex]\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Where \( a = 1 \), \( b = -4.25 \), and \( c = 1 \).
5. Calculate the discriminant:
First, find the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-4.25)^2 - 4 \cdot 1 \cdot 1 = 18.0625 - 4 = 14.0625 \][/tex]
6. Find the roots:
Now plug the discriminant (\(\Delta\)) into the quadratic formula:
[tex]\[y_1 = \frac{4.25 + \sqrt{14.0625}}{2} = \frac{4.25 + 3.75}{2} = \frac{8}{2} = 4\][/tex]
[tex]\[y_2 = \frac{4.25 - \sqrt{14.0625}}{2} = \frac{4.25 - 3.75}{2} = \frac{0.5}{2} = 0.25\][/tex]
So the solutions for \( y \) are 4 and 0.25.
7. Revert the substitution:
Recall that \( y = 2^x \). Therefore, we have:
[tex]\[2^x = 4\][/tex]
[tex]\[2^x = 0.25\][/tex]
8. Solve for \( x \):
To find \( x \), we need to express each equation in terms of \( x \) using logarithms:
[tex]\[2^x = 4 \implies x = \log_2 4 = 2\][/tex]
[tex]\[2^x = 0.25 \implies x = \log_2 0.25 = -2\][/tex]
Thus, the solutions for \( x \) are \( x = 2 \) and \( x = -2 \).
In conclusion, the solution to the equation \(2^x + \frac{1}{2^x} = 4.25\) is:
[tex]\[ x = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
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